Question

Solve the homogeneous DE with n = 2, (x^2 - 3y^2) dx + 2xy dy = 0 when v = y/x

Answer 1

The way I go about it I end up with lnx - (1/6) ln(1-3(y/x)^2) = c
the book's answer is y = x sqrt(x+1) and y = -(x sqrt(x+1))

Answer 2

\(v\) is not supposed to be an integration factor is it?

Answer 3

no, the book says to substitute and then convert into a variable separable problem

Answer 4

I understand, lets work both on this then. Maybe we will figure out something together.

Answer 5

First I want to rearrange a bit, I understand now your books substitution because the DE given above holds with following statement:

\[\Large f(x,y)=f(tx,ty) \]
Or in different words, scalar multiplication doesn't change the equation at all.

Answer 6

So after rearranging I get:
\[\Large \left(1-3\left(\frac{y}{x}\right)^2\right)+2\left(\frac{y}{x}\right) \frac{dy}{dx}=0 \]
For \( x \neq 0 \)

Answer 7

what I did was solve \[v = y/x \] for y and got \[y = vx\]
from there you put it into the equation
\[(x^2 - 3x^2v^2) dx + 2(x^3v^2)(v dx+ xdv)\]

Answer 8

Now lets perform the substitution:
\[\Large z=\frac{y}{x} \]
Or direct
\[\Large y=zx \]

so
\[\Large \frac{dy}{dx}=\frac{dz}{dx}x+z \]

Answer 9

Now we can back substitute:
\[\Large (1-3z^2)+2z\left(\frac{dz}{dx}x+z\right)=0 \]

Or \[\Large 1-z^2+2zx\frac{dz}{dx}=0 \]

Answer 10

Do you see how to continue from here? Or any disagreement with my steps? I would now try to separate the given DE.

Answer 11

I'm confused as to how you got the first equation your replied with

Answer 12

I divided the whole equation first with x^2 and after with dx

Answer 13

if I didn't make any careless mistakes, that's what you should get from it

Answer 14

If you find any mistakes I have made, feel free to correct me.

Answer 15

I don't see any mistakes, it's just that the way you're doing it is not the way I'm accustomed to so working through it is a bit difficult

Answer 16

okay, basically, all I am trying to do is setup the original DE so that the substitution makes more sense to me (optically). Because in the problem it says, that we should use

\[\Large v=\frac{y}{x}
\]
I used \(z\) instead of \(v\) because I am more used to it. Anyway, since they give us this help, I wanted the original DE to include as many terms such as \(v\) as possible, I obtained this by dividing the equation through \(x^2\) it gives me \(v^2\) term and a \(v\) term.

I also rewrote the substitution given above, so it's easier for me to differentiate:

\[\Large v=\frac{y}{x} \longrightarrow y=vx \]

This way it's much easier for me to differentiate, using the product rule.

Answer 17

is it possible to just end up with a different equation than what the book says because no matter what I do I still get the same answer

Answer 18

Well the following differential equation I obtained above:
\[\Large 1-z^2+2zx\frac{dz}{dx}=0 \]
Can be separated, this gives you one differential equation.

Answer 19

would it be
\[dz/(1-z) = dx/-2x\]?

Answer 20

Lets rearrange the equation together:
\[\Large z^2-1=2zx\frac{dz}{dx} \]
Divide with z

\[\Large \frac{z^2-1}{z}=2x\frac{dz}{dx} \]

Compute the inverse

\[\Large \frac{z}{z^2-1}=\frac{1}{2x}\cdot \frac{dx}{dz} \]

Answer 21

so, \[1/2\ln (z^2-1) = lnx/2 + c\]

Answer 22

now you need to back substitute and solve for Y if possible.

Answer 23

I ended up with, \[y = x \sqrt{c\frac{ x }{ 2 }+1}\]
not sure if that's correct

Answer 24

Seems like the 2 cancels out and gets absorbed by the constant:

\[\Large z^2-1=\exp(x+C) \\
\Large \left(\frac{y}{x}\right)^2=Cx+1 \]

So I think:
\[\Large y=\pm x\sqrt{Cx+1} \]

Answer 25

thank you so much for your help, you're a life saver